Question: Simplify the following expression: $\dfrac{4y}{48y^3}$ You can assume $y \neq 0$.
Solution: $ \dfrac{4y}{48y^3} = \dfrac{4}{48} \cdot \dfrac{y}{y^3} $ To simplify $\frac{4}{48}$ , find the greatest common factor (GCD) of $4$ and $48$ $4 = 2 \cdot 2$ $48 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$ $ \mbox{GCD}(4, 48) = 2 \cdot 2 = 4 $ $ \dfrac{4}{48} \cdot \dfrac{y}{y^3} = \dfrac{4 \cdot 1}{4 \cdot 12} \cdot \dfrac{y}{y^3} $ $\phantom{ \dfrac{4}{48} \cdot \dfrac{1}{3}} = \dfrac{1}{12} \cdot \dfrac{y}{y^3} $ $ \dfrac{y}{y^3} = \dfrac{y}{y \cdot y \cdot y} = \dfrac{1}{y^2} $ $ \dfrac{1}{12} \cdot \dfrac{1}{y^2} = \dfrac{1}{12y^2} $